Question

How do you solve ${{x}^{2}}+{{y}^{2}}=20$ and $x+y=6$ ?

Answer 1

Hint: Here in this question we have been asked to solve the two given algebraic expressions ${{x}^{2}}+{{y}^{2}}=20$ and $x+y=6$ . Now we will simplify any one expression for example $x+y=6$ and write it as $y=6-x$ and substitute it in the expression ${{x}^{2}}+{{y}^{2}}=20$.

Complete step-by-step solution:
Now considering from the question we have been asked to solve the two given algebraic expressions ${{x}^{2}}+{{y}^{2}}=20$ and $x+y=6$ .
Now we will simplify any one expression for example $x+y=6$ and write it as $y=6-x$ and substitute it in the expression ${{x}^{2}}+{{y}^{2}}=20$ .
After doing that we will have ${{x}^{2}}+{{\left( 6-x \right)}^{2}}=20$ . Now we will expand this expression after doing that we will have $\Rightarrow {{x}^{2}}+36+{{x}^{2}}-12x=20$ .
Now we will further simplify the expression $\Rightarrow 2{{x}^{2}}-12x+16=0$ .
Here we have a quadratic expression in the form of $a{{x}^{2}}+bx+c=0$ , we know that the roots of this expression are given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
By using this formula we will simplify the expression $2{{x}^{2}}-12x+16=0$ by doing this we will have $\Rightarrow \dfrac{12\pm \sqrt{{{12}^{2}}-4\left( 2 \right)\left( 16 \right)}}{2\left( 2 \right)}$ .
Now by simplifying this we will have
$\begin{align}
  & \Rightarrow \dfrac{12\pm \sqrt{144-\left( 8 \right)\left( 16 \right)}}{4}=\dfrac{12\pm \sqrt{144-128}}{4} \\
 & \Rightarrow \dfrac{12\pm \sqrt{16}}{4}=\dfrac{12\pm 4}{4} \\
 & \Rightarrow \dfrac{16}{4},\dfrac{8}{4}=4,2 \\
\end{align}$
Therefore we can conclude that the solutions of the given algebraic expressions ${{x}^{2}}+{{y}^{2}}=20$ and $x+y=6$ are $4,2$.

Note: While answering questions of this type we should be sure with the concepts that we apply in the process and the calculations that we perform in between the steps. This is a very simple and easy question and can be answered accurately in a short span of time. Alternatively we can write the first expression as $x=6-y$ and substitute it in the expression and write it as ${{\left( 6-y \right)}^{2}}+{{y}^{2}}=20$ . Then we will get the solutions as $4,2$ by using the same process we used before.